Re: Help with recursion

In article <1138730402.129977.241530@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Chip Eastham <hardmath@xxxxxxxxx> wrote:
>
>Robert Israel wrote:
>> In article <1138694385.335293.29860@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
>> robert.w.adams@xxxxxxxxxxx <robert.w.adams@xxxxxxxxxxx> wrote:
>> >OK, if I do that then I get
>> >
>> >sum(2 to n) x_sub_n = sum(2 to n) x_sub(n-1) + sum(2 to n) K/log(n)
>> >
>> >and if I subtract sum(2 to n) x_sub(n-1) from both sides;
>> >
>> >x_sub_n = sum(2 to n) K/log(n)
>> >
>> >Is there a closed-form solution for sum(2 to n) K/log(n) ?
>>
>> Don't use the same letter for the dummy variable as for an endpoint
>> of the summation. You mean sum_{j=2}^n K/log(j). No, there isn't
>> a closed-form solution AFAIK.
>
>The factor K comes out of the summation, of course, and I would
>conjecture that, analogous with the Euler-Mascharoni constant,
>for large n, SUM 1/ln(j) FOR j = 2 to n minus the logarithmic
>integral li(n) = INTEGRAL 1/ln(x) dx FROM x = 2 to n approaches
>some fixed constant. Purely speculation.

True. Look up "Euler-Maclaurin summation formula" for more.

Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada


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